InterviewSolution
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InterviewSolution
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Determine which of the following polynomials has x – 1 factor.(i) x4 – 2x3 – 3x2 + 2x + 2(ii) x4 + x3 + x2 + x + 1(iii) x4 + 3x3 – 3x2 + x – 2(iv) x3 – x2 – (2 + √3)x + √3 |
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Answer» (i) Let p(x) = x4 – 2x3 – 3x2 + 2x + 2 and If x – 1 i.e. g(x) be a factor of p(x) then P(1) = 0 Now p(1) = (1)4 – 2(1)3 – 3(1)2 + 2(1) + 2 = 1 – 2 – 3 + 2 + 2 = 0 ∵ p(1) = 0 => x – 1 is a factor of p(x) = x4 – 2x3 – 3x2 + 2x + 2. (ii) Let p(x) = x4 + x3 + x2 + x + 1 If x – 1 is a factor of p(x) then p(1) = 0 p(1) = (1)4 + (1)3 + (1)2 + (1) + 1 = 1 + 1 + 1 + 1 + 1 = 5 p(1) ≠ 0 ∴ x – 1 is a factor of x4 + x3 + x2 + x + 1. (iii) Let p(x) = x4 + 3x3 – 3x2 + x – 2 If x – 1 is a factor of p(x) then p(1) = 0 ∴ p(1) = (1)4 + 3(1)3 – 3(1)2 + (1) – 2 = 1 + 3 – 3 + 1 – 2 = 0 p(1) = 0 ∴ x – 1 is a factor of x4 + 3x3 – 3x2 + x – 2. (iv) Let p(x) = x3 – x2 – (2 + √3)x + √3 If x – 1 is a factor of p(x) then p(1) = 0 ∴ p(1)= (1)3 – (1)2 – (2 + √3)(1) + √3 = 1 – 1 – 2 – √3 + √3 = -2 ∴ p( 1) ≠ 0 Hence, x – 1 is not a factor of x3 – x2 – (2 + √3)x + √3 |
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